165
7
Beam and Frame Analysis:
Displacement Method—Part I
the.best.computing.tool.was.a.slide.rule,.to.solve.frame.problems.that.nor- mally.require.the.solution.of.simultaneous.algebraic.equations..Its.relevance.
today,.in.the.era.of.the.personal.computer,.is.in.its.insight.on.how.a.beam.and.
frame.react.to.applied.loads.by.rotating.its.nodes.and.thus.distributing.the.
loads.in.the.form.of.member-end.moments.(MEMs)..Such.an.insight.is.the.
foundation.of.the.modern.displacement.method.
Take.the.very.simple.frame.in.the.following.figure.as.an.example..The.
externally.applied.moment.at.node.b.tends.to.create.a.rotation.at.node.b..
Because.member.ab.and.member.bc.are.rigidly.connected.at.node.b,.the.
same.rotation.must.take.place.at.the.end.of.member.ab.and.member.bc..
For.rotation.at.the.end.of.member.ab.and.member.bc.to.happen,.an.end.
moment.must.be.internally.applied.at.the.member.end..This.member-end.
moment.comes.from.the.externally.applied.moment..Nodal.equilibrium.
at.b. requires. the. applied. external. moment. of. 100. kN. be. distributed. to.
the.two.ends.of.the.two.joining.members.at.b..How.much.each.member.
will.receive.depends.on.how.“rigid”.each.member.is.in.its.resistance.to.
rotation.at.b..Since.the.two.members.are.identical.in.length,.L,.and.cross- section. rigidity,.EI,. we. assume. for. the. time. being. that. they. are. equally.
rigid..Thus,.half.of.the.100.kN-m.goes.to.member.ab.and.the.other.half.
goes.to.member.bc.
100 kN-m
EI, L a
c b
100 kN-m
50 kN-m 50 kN-m
b EI, L EI, L
EI, L
a
c b b
50 kN-m 50 kN-m
Moment equilibrium of node b
A.frame.example.showing.member-end.moments.
In. the. preceding. figure,. only. the. member-end. moments. are. shown.. The.
member-end.shear.and.axial.forces.are.not.shown.to.avoid.overcrowding.the.
figure..The.distributed.moments.(DMs).are.“member-end”.moments.denoted.
by.Mba.and.Mbc,.respectively..The.sign.convention.of.member-end.moments.and.
applied.external.moments.is.clockwise.is.positive..We.assume.the.two.members.
are.equally.rigid.and.receive.half.of.the.applied.moment,.not.only.because.they.
appear.to.be.equally.rigid.but.also.because.each.of.the.two.members.is.under.
identical.loading.conditions:.fixed.at.the.far.end.and.hinged.at.the.near.end.
In.other.cases,.the.beam.and.column.may.not.be.of.the.same.rigidity,.but.
they.may.have.the.same.loading.and.supporting.conditions:.fixed.at.the.far.
end.and.allowed.to.rotate.at.the.near.end..This.configuration.is.the.funda- mental.configuration.of.moment.loading.from.which.all.other.configurations.
can.be.derived.by.the.principle.of.superposition..We.shall.delay.the.deriva- tion.of.the.governing.formulas.until.we.have.learned.the.operating.proce- dures.of.the.moment.distribution.method.
Mba EI, L
a
c b
b Mbc
θb θb
Beam.and.column.in.a.fundamental.configuration.of.a.moment.applied.at.the.end.
Suffice.it.to.say.that.given.the.loading.and.support.conditions.shown.in.the.
following.figure,.the.rotation.θb.and.the.member-end.moment.Mba.at.the.near.
end,.b,.are.proportional..The.relationship.between.Mba and.θb.is.expressed.in.
the.following.equation,.the.derivation.of.which.will.be.given.later.
Mba = 4EKθb EI, L
a θb b
Mba = 2EKθb
Vba= 6EKθb/L Vab = 6EKθb/L
The.fundamental.case.and.the.reaction.solutions.
Mba =.4(EK)abθb. (7.1a) where.Kab =( / )I Lab.
We.can.write.a.similar.equation.for.Mbc.of.member.bc.
Mbc =.4(EK)bcθb. (7.1b) where.Kbc=( / )I Lbc.
Furthermore,.the.moment.at.the.far.end.of.member.ab,.Mab at.a.is.related.to.
the.amount.of.rotation.at.b.by.the.following.formula:
Mab =.2(EK)abθb. (7.2a) Similarly,.for.member.bc,
Mcb =.2(EK)bcθb. (7.2b) As.a.result,.the.member-end.moment.at.the.far.end.is.one-half.of.the.near- end.moment:
M = 1M
ab 2 ba. (7.3a)
and
M =1M
cb 2 bc. (7.3b)
Note.that.in.the.preceding.equations,.it.is.important.to.keep.the.subscripts.
because.each.member.may.have.a.different.EK.
The. significance. of. Equation. 7.1. is. that. it. shows. that. the. amount. of. the.
member-end. moment,. distributed. from. the. unbalanced. nodal. moment,. is.
proportional.to.the.member.stiffness.4EK, which.is.the.moment.needed.at.
the.near.end.to.create.a.unit.rotation.at.the.near.end,.while.the.far.end.is.
fixed..Consequently,.when.we.distribute.the.unbalanced.moment,.we.need.
only.to.know.the.relative.stiffness.of.each.of.the.joining.members.at.that.par- ticular.end..The.equilibrium.equation.for.moment.at.node.b.is
Mba +.Mbc =.100.kN-m. (7.4) Since
Mba.:.Mbc =.(EK)ab :.(EK)bc
we.can.“normalize”.the.previous.equation.so.that.both.sides.would.add.up.
to.one,.that.is.100%,.utilizing.the.fact.that.(EK)ab = (EK)bc.in.the.present.case:
M M M
M M M
EK EK EK
EK EK EK
: ( )
( ) ( ) : ( )
( ) ( )
1 2:1
2
ba
ba bc
bc
ba bc
ab
ab bc
bc
ab bc
+ + =
+ + =
... (7.5)
Consequently
. M 1 M M
2( ) 1
2(100 kN-m) 50 kN-m
ba= ba+ bc = =
. M 1 M M
2( ) 1
2(100 kN-m) 50 kN-m
bc= ba+ bc = =
From.Equation.7.2,.we.obtain M 1M
2 25 kN-m
ab= ba =
. M 1M
2 25 kN-m
cb= bc =
Now. that. all. the. member-end. moments. are. obtained,. we. can. proceed.
to.find.member-end.shears.and.axial.forces.using.the.free-body.diagrams.
(FBDs).shown.next.
Vab Vba
Vbc
Vcb 25 kN-m
25 kN-m
50 kN-m 50 kN-m
FBDs.to.find.shear.and.axial.forces.
The.dashed.lines.indicate.that.the.axial.force.of.one.member.is.related.to.
the. shear. force. from. the. joining. member. at. the. common. node.. The. shear.
forces.are.computed.from.the.equilibrium.conditions.of.the.FBDs:
.. V =V = M +M
ab ba baL ab
ab
and
. .V =V = M +M
bc cb bcL cb
bc
The.moment.and.deflection.diagrams.of.the.whole.structure.are.shown.next.
50 kN-m 25 kN-m
25 kN-m +
M ∆
Moment.and.deflection.diagrams.
In.drawing.the.moment.diagram,.note.that.the.sign.conventions.for.inter- nal. moment. (as. in. moment. diagram). and. the. member-end. moment. (as. in.
Equation.7.1.through.Equation.7.5).are.different..The.former.depends.on.the.
orientation.and.which.face.the.moment.is.acting.on,.and.the.latter.depends.
only.on.the.moment.direction.(clockwise.is.positive).
Positive internal moment
Positive member-end moment Positive internal moment Negative member-end moment Difference.in.sign.conventions.
Let.us.recap.the.operational.procedures.of.the.moment.distribution.method:
. 1..Identify. the. node. that. is. free. to. rotate.. In. the. present. case,. it. was.
node.b..The.number.of.“free”.rotating.nodes.is.called.the.degree.of.
freedom.(DOF)..In.the.present.case,.the.DOF.is.one.
. 2..Identify.the.joining.members.at.this.node.and.compute.their.rela- tive.stiffness.according.to.Equation.7.5,.which.can.be.generalized.to.
cover.more.than.two.members.
M M
M M
M M
EK EK
EK EK
EK
: : ( ) EK
( ) : ( )
( ) : ( )
( )
ab xy
bc xy
cd xy
ab xy
bc xy
cd
∑ ∑ ∑ … = xy
∑ ∑ ∑ …
. . where.the.summation.is.over.all.joining.members.at.the.particular.
node..Each.of.the.expressions.in.this.equation.is.called.a.distribution.
factor.(DF),.which.adds.up.to.1.or.100%..Each.of.the.moment.at.the.
end.of.a.member.is.called.a.member-end.moment.
. 3..Identify.the.unbalanced.moment.at.this.node..In.the.present.case,.it.
was.100.kN-m.
. 4..To.balance.the.100.kN-m,.we.need.to.add.–100.kN-m.to.the.node,.
which,.when.viewed.from.the.member.end,.becomes.positive.100.kN-m..
This.100.kN-m.is.distributed.to.member.ab.and.bc.according.to.the.
DF.of.each.member..In.this.case.the.DF.is.50%.each..Consequently,.
50.kN-m.goes.to.Mba.and.50.kN-m.goes.to.Mbc..They.are.called.the.
distributed.moment..Note.that.the.externally.applied.moment.is.dis- tributed.as.member-end.moments.in.the.same.sign,.that.is,.positive.
to.positive.
. 5..Once.the.balancing.moment.is.distributed,.the.far.ends.of.the.joining.
members.should.receive.50%.of.the.distributed.moment.at.the.near.
end..The.factor.of.50%.or.½.is.called.the.carryover factor.(COF)..The.
moment.at.the.far.end.thus.distributed.is.called.the.carryover moment.
(COM)..In.the.present.case,.they.are.25.kN-m.for.Mab.and.25.kN-m.
for.Mcb,.respectively.
. 6..We.note.that.at.the.two.fixed.ends,.whatever.moments.are.carried.
over,. they. are. balanced. by. the. support. reaction.. That. means. the.
moment.equilibrium.is.achieved.at.the.fixed.ends.with.no.need.for.
additional. distribution.. This. is. equivalent. to. say. that. the. stiffness.
of.the.support.relative.to.the.stiffness.of.the.member.is.infinite..Or,.
even.simpler,.we.may.formally.designate.the.distribution.factors.at.
a.fixed.support.as.1:0,.with.one.being.assigned.for.the.support.and.
zero.assigned.to.the.member..The.zero.DF.means.we.need.not.redis- tribute.any.moment.at.the.member.end.
. 7..The.moment.distribution.method.operations.end.when.all.the.nodes.
are.in.moment.equilibrium..In.the.present.case,.node.b.is.the.only.
node. we. need. to. concentrate. on. and. it. is. in. equilibrium. after. the.
unbalanced.moment.is.distributed.
. 8..To.complete.the.solution.process,.however,.we.still.need.to.find.the.
other.unknowns.such.as.shear.and.axial.forces.at.the.end.of.each.
member..That.is.accomplished.by.drawing.the.FBD.of.each.member.
and.writing.equilibrium.equations.
. 9..The.moment.diagram.and.deflection.diagram.can.then.be.drawn.
We.shall.now.go.through.the.solution.process.by.solving.a.similar.problem.
with.a.single.degree.of.freedom.(SDOF).
Example 7.1
Find all the member-end moments of the beam shown. EI is constant for all members.
30 kN-m
a b c
10 m 5 m
Beam problem with a SDOF.
Solution
1. Preparation.
a. Unbalanced moment: At node b there is an externally applied moment (EAM), which should be distributed as member-end moments in the same sign.
b. The distribution factors at node b:
.
EK EK EI
L
EI
ba bc ab bc L
ab bc
DF : DF 4 : 4 4 : 4 1
10:1
5 0.33 : 0.67
= = = =
c. As a formality, we also include DFab = 0, and DFbc = 0, at a and c, respectively.
2. Tabulation. All the computing can be tabulated as shown next.
The arrows indicate the destination of the carryover moment. The dashed lines show how the DF is used to compute the DM.
Moment Distribution Table for an SDOF Problem
Mab Mba Mbc Mcb
Node a b c
Member ab bc
DF 0 0.33 0.67 0
MEM1
EAM2 30
DM3 +10 +20
COM4 +5 +10
Sum5 +5 +10 +20 +10
1. Member-end moment.
2. Externally applied moment.
3. Distributed member-end moment.
4. Carryover moment.
5. Sum of member-end moments.
3. Post moment-distribution operations. The moment and deflection diagrams are shown next.
20 kN-m
–10 kN-m –10 kN-m
5 kN-m
Inflection Point
Moment and deflection diagrams.
The.moment.distribution.method.becomes.iterative.when.there.are.more.
than.one.DOF..The.aforementioned.procedures.for.one.DOF.problem.can.
still.apply.if.we.consider.one.DOF.at.a.time..That.is.to.say.that.when.we.
concentrate. on. one. DOF,. the. other. DOFs. are. considered. “locked”. into. a.
fixed. support. and. are. not. allowed. to. rotate.. When. the. free. node. gets. its.
distributed. moment. and. the. carryover. moment. reaches. the. neighboring.
and.previously.locked.node,.that.node.becomes.unbalanced,.thus.requir- ing.“unlocking”.to.distribute.the.balancing.moment,.which.in.turn.creates.
carryover.moment.at.the.first.node..That.requires.another.round.of.distribu- tion.and.carrying.over..Thus.begins.the.cycle.of.“locking–unlocking”.and.
the.balancing.of.moments.from.one.node.to.another..We.shall.see,.however,.
in.each.subsequent.iteration,.the.amount.of.unbalanced.moment.becomes.
progressively. smaller.. The. iteration. stops. when. the. unbalanced. moment.
becomes. negligible.. This. iterative. process. is. illustrated. in. the. following.
example.of.two.DOFs.
Example 7.2
Find all the member-end moments of the beam shown. EI is constant for all members.
3 m 5 m 5 m
a b c d
30 kN-m
Example of a beam with two DOFs.
Solution
1. Preparation.
a. Both nodes b and c are free to rotate. We choose to balance node c first.
b. Compute DF at b:
= =
EK EK EI
L
EI
ba bc ab bc L
ab bc
DF : DF = 4 : 4 = 4 : 4 1
3:1
5 0.625 : 0.375 c. Compute DF at c:
= EK EK = EI = =
L
EI
cb cd bc cd L
bc cd
DF : DF 4 : 4 4 : 4 1
5:1
5 0.5 : 0.5 d. Assign DF at a and d: DFs are zero at a and d.
2. Tabulation.
Moment Distribution for a Two-DOF Problem
Node a b c d
Member ab bc cd
DF 0 0.625 0.375 0.5 0.5 0
MEM Mab Mba Mbc Mcb Mcd Mdc
EAM 30
DM +15 +15
COM +7.50 +7.50
DM –4.69 –2.81
COM –2.35 –1.41
DM +0.71 +0.70
COM +0.36 0.35
DM –0.22 –0.14
COM –0.11 –0.07
DM +0.04 +0.03
COM +0.02 +0.02
DM –0.01 –0.01
COM 0.00 0.00
Sum –2.46 –4.92 +4.92 +14.27 +15.73 +7.87 In the table, the encircled moment is the unbalanced moment.
Note how the circles move back and forth between nodes b and c. Also note how the EAM at c and the unbalanced moment, cre- ated by the COM at b, are treated differently. The EAM is balanced by distributing the amount in the same sign to the member ends, while the unbalanced moment at a node is balanced by distribut- ing the negative of the unbalanced moment to the moment ends.
3. Post moment-distribution operations. The moment and deflection diagrams are shown next.
–2.46 4.92
–14.27 15.73
–7.87
Inflection Point
Moment and deflection diagrams.
Treatment of load between nodes. In.the.previous.examples,.the.applied.load.
was. an. applied. moment. at. a. node.. We. can. begin. the. distribution. process.
right.at.the.node..In.most.practical.cases,.the.load.will.be.either.concentrated.
loads. or. distributed. loads. applied. between. nodes.. These. cases. call. for. an.
additional.step.before.we.can.begin.the.distribution.of.moments.
Load.applied.between.nodes.
We.imagine.that.all.the.nodes.are.“locked”.at.the.beginning..Then.each.
member. is. in. a. state. of. a. clamped. beam. with. a. transverse. load. applied.
between.the.two.ends.
a b
MFba
MFab
P
Fixed-end.beam.with.applied.load.
The.moment.needed.to.“lock”.the.two.ends.are.called.fixed-end.moments.
(FEMs)..They.are.positive.if.acting.clockwise..For.typical.loads,.the.FEMs.can.
be.precomputed.and.are.tabulated.in.the.FEM.table.given.at.the.end.of.this.
chapter..These.FEMs.are.to.be.balanced.when.the.node.is.“unlocked”.and.
allowed. to. rotate.. Thus,. the. effect. of. the. transverse. load. applied. between.
nodes.is.to.create.moments.at.both.ends.of.a.member..These.FEMs.should.be.
balanced.by.moment.distribution.
Example 7.3
Find all the member-end moments of the beam shown. EI is constant for all members.
2 m 2 m 4 m
a c
b 3 kN/m 4 kN
Example with load applied between nodes.
Solution
1. Preparation.
a. Only node b is free to rotate. There is no externally applied moment at node b to balance, but the transverse load between nodes creates FEMs.
b. FEM for member ab. The concentrated load of 4 kN creates FEMs at end a and end b. The formula for a single transverse load in the FEM table gives us:
. M P Length
abF ( ) ( ) 8
(4) (4)
8 2kN-m
= − = − = −
. M P Length
baF ( ) ( ) 8
(4) (4)
8 2kN-m
= = =
c. FEM for member bc. The distributed load of 3 kN/m creates FEMs at end b and end c. The formula for a distributed trans- verse load in the FEM table gives us:
. M w Length
bcF ( ) ( ) 12
(3) (4)
12 4 kN-m
2 2
= − = − = −
. M w Length
cbF ( ) ( ) 12
(3) (4)
12 4 kN-m
2 2
= = =
d. Compute DF at b:
.
= EK EK EI = =
L
EI
ba bc ab bc L
ab bc
DF : DF 4 : 4 = 4 : 4 1
4:1
4 0.5: 0.5 e. Assign DF at a and c: DFs are zero at a and c.
2. Tabulation.
Moment Distribution for an SDOF Problem with FEMs
Node a b c
Member ab bc
DF 0 0.5 0.5 0
EAMMEM Mab Mba Mbc Mcb
FEM –2 +2 –4 +4
DM +1 +1
COM +0.5 +0.5
Sum –1.5 +3 –3 +4.5
3. Post moment-distribution operations. The shear forces at both ends of a member are computed from the FBDs of each member. Knowing the member-end shear forces, the moment diagram can then be drawn. The moment and deflection diagrams are shown next.
FBDs of the two members.
–1.5 –3
–4.5
1.75 5.79
1.875 m
Inflection Point
1.5 kN-m 2 m 2 m
4 kN
3 kN-m 2.38 kN 1.62 kN
3 kN-m 4 m
3 kN/m
4.5 kN-m 6.38 kN 5.62 kN
Moment and deflection diagrams.
Treatment of hinged ends.. At. a. hinged. end,. the. MEM. is. equal. to. zero. or.
whatever.an.externally.applied.moment.is.at.the.end..During.the.process.of.
moment.distribution,.the.hinged.end.may.receive.COM.from.the.neighbor- ing.node..That.COM.must.then.be.balanced.by.distributing.100%.of.it.at.the.
hinged.end..This.is.because.the.distribution.factor.of.a.hinged.end.is.1.or.
100%;.the.hinged.end.may.be.considered.to.be.connected.to.air,.which.has.
zero.stiffness..This.new.distributed.moment.starts.another.cycle.of.carryover.
and.distribution..This.process.is.illustrated.in.Example.7.4.
The.cycle.of.iteration.is.greatly.simplified.if.we.recognize.at.the.very.begin- ning.of.moment.distribution.that.the.stiffness.of.a.member.with.a.hinged.
end. is. fundamentally. different. from. that. of. the. standard. model. with. the.
far.end.fixed..We.will.delay.the.derivation.but.will.state.that.the.moment.
needed.at.the.near.end.to.create.a.unit.rotation.at.the.near.end.with.the.far.
end.hinged.is.3EK,.less.than.the.4EK.if.the.far.end.is.fixed.
Mba = 3EKθb θb
θb
Mba = 4EKθb Mab= 2EKθb
Mba = 0 a a
b
b
Member.with.a.hinged.end.versus.the.standard.model.with.the.far.end.fixed.
Note.that.there.is.no.COM.at.the.hinged.end.(Mba.=.0).if.we.take.the.mem- ber.stiffness.factor.as.3EK.instead.of.4EK..We.can.thus.compute.the.relative.
distribution.factors.accordingly,.and.when.distributing.the.moment.at.one.
end.of.the.member,.we.need.not.carry.over.the.distributed.moment.to.the.
hinged.end..This.simplified.process.with.a.modified.stiffness.from.4EK.to.
3EK.is.illustrated.in.Example.7.5.
Example 7.4
Find all the member-end moments of the beam shown. EI is constant for all members.
2 m 2 m 4 m
a c
b 3 kN/m 4 kN
2 m 2 kN
2 m 2 m 4 m
a c
b 3 kN/m
4 kN-m 4 kN
Turning a problem with a cantilever end into one with a hinged end.
Solution
The original problem with a cantilever end can be treated as one with a hinged end as shown. We shall solve only the problem with a hinged end.
Note that the vertical load is not shown in the equivalent hinged-end prob- lem because it is taken up by the support at a.
1. Preparation. Since the geometry and loading are similar to that of Example 7.3, we can copy the preparation part but note that an externally applied moment is present.
a. Only nodes b and a are free to rotate. There is an externally applied moment at node a and the transverse load between nodes create FEMs at all nodes.
b. FEM for member ab. The concentrated load of 4 kN creates FEMs at end a and end b. The formula for a single transverse load in the FEM table gives us:
. M P Length
abF ( ) ( ) 8
(4)(4)
8 2kN-m
= − = − = −
M P Length
baF ( ) ( ) 8
(4)(4)
8 2kN-m
= = =
c. FEM for member bc. The distributed load of 3 kN/m creates FEMs at end b and end c. The formula for a distributed trans- verse load in the FEM table gives us:
. M w Length
bcF ( ) ( ) 12
(3) (4)
12 4kN-m
2 2
= − = − = −
. M w Length
cbF ( ) ( ) 12
(3) (4)
12 4 kN-m
2 2
= = =
d. Compute DF at b:
.
= EK EK = EI = =
L
EI
ba bc ab bc L
ab bc
DF : DF 4 : 4 4 : 4 1
4:1
4 0.5: 0.5 e. Assign DF at a and c: DFs are one at a and zero at c.
2. Tabulation. In the moment distribution process shown next, we must deal with the unbalanced moment at the hinged end first.
The EAM of –4 kN-m and the FEM of –2 kN-m at node a add up to 2 kN-m of unbalanced moment, not –6 kN-m. This is because the FEM and DM at node a should add up to the EAM, which is
–4 kN-m. Thus, we need to distribute (–4 kN-m) – (–2 kN-m) = –2 kN-m to make the node balanced. The formula to remember is DM = EAM – FEM. This formula is applicable to all nodes where there are both EAMs and FEMs.
Moment Distribution Table for a Beam with a Hinged End
Node a b c
Member ab bc
DF 1 0.5 0.5 0
MEM Mab Mba Mbc Mcb
EAM –4
FEM –2 +2 –4 +4
DM –2
COM –1
DM +1.5 +1.5
COM +0.8 +0.8
DM –0.8
COM –0.4
DM +0.2 +0.2
COM +0.1 +0.1
DM –0.1
COM 0.0
Sum –4 +2.3 –2.3 +4.9
The. aforementioned. back-and-forth. iteration. between. nodes.a. and.b. is.
avoided.if.we.use.the.simplified.procedures.as.illustrated.next.
Example 7.5
Find all the member-end moments of the beam shown. EI is constant for all members. Use the modified stiffness to account for the hinged end at node a.
2 m 2 m 4 m
a c
b 3 kN/m 4 kN-m 4 kN
Beam with a hinged end.
Solution
1. Preparation. Note the stiffness computation in step d.
a. Only nodes b and a are free to rotate. Node a is considered a hinged node and needs no moment distribution except at the very beginning. There is an externally applied moment at node a and the transverse load between nodes create FEMs at all nodes.